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In this answer

https://math.stackexchange.com/a/2175371/829738

more precisely in this section

"Done that, now prove that $[0, 1]/∼$ is homeomorphic to $\mathbb R/ \mathbb Z$. For that, consider $f:[0, 1]→ \mathbb R$ the inclusion, compose with the quotient to yield a function $[0,1]→\mathbb R/ \mathbb Z$, then use the universal property to get a function $[0,1]/∼→\mathbb R/ \mathbb Z$, which is a continuous bijection from a compact set to a Hausdorff set again."

the map $[0, 1] \rightarrow \mathbb R/ \mathbb Z$ is a quotient map for what reason? Compact-Hausdorff?

PaulichenT
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There is no claim that the composite map from $[0,1]$ to $\Bbb R/\Bbb Z$ is a quotient map; what matters is that it’s continuous. The goal is to show that $\Bbb R/\Bbb Z$ is compact.

Suppose that $f:X\to Y$ is a continuous bijection from a compact space to a Hausdorff space. If $K\subseteq X$ is closed, then $K$ is compact, so continuity of $f$ ensures that $f[K]$ is compact, and Hausdorffness of $Y$ then ensures that $f[K]$ is closed in $Y$. Thus, $f$ is a closed, continuous bijection and is therefore a homeomorphism, so $\Bbb R/\Bbb Z$ is compact.

Brian M. Scott
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  • Nice! I was really lost. Thank you – PaulichenT Oct 23 '20 at 03:28
  • @PaulichenT: You’re welcome. – Brian M. Scott Oct 23 '20 at 03:33
  • Just one more question. The map $g: [0, 1] /\sim \rightarrow \mathbb R / \mathbb Z$ is a continuous bijection, since in addition to $\iota$ and $ \pi_{\mathbb R / \mathbb Z}$ being continuous, $\pi_{\mathbb R/\mathbb Z} \circ \iota$ satisfies \begin{align} \pi_{\mathbb R / \mathbb Z} \circ \iota(t)=\pi_{\mathbb R/\mathbb Z}\circ \iota(t ') \Leftrightarrow t \sim t', \end{align} right? However, it turns out that in the argument you mentioned, continuous bijection must be compact for Hausdorff. I probably don't understand any arguments, sorry I don't have much experience in this area. – PaulichenT Oct 23 '20 at 15:00
  • @PaulichenT: I’m not sure exactly what your question is here; can you clarify a bit? – Brian M. Scott Oct 23 '20 at 15:33
  • When using the universal property, an application $[0, 1]/ \sim \rightarrow \mathbb R/ \mathbb Z$ is obtained, this being a continuous bijection, but in this case this application it goes from a Hausdorff to a compact, and not from a compact to a Hausdorff, is that a problem? – PaulichenT Oct 23 '20 at 15:50
  • @PaulichenT: That map isn’t a bijection; it is a continuous function from $[0,1]$ to $\Bbb R/\Bbb Z$ that respects the equivalence relation $\sim$, i.e., takes the same value at $0$ and $1$. That’s all that is needed to use the universal property to get a continuous bijection from $[0,1]/\sim$ to $\Bbb R/\Bbb Z$. We already know that $[0,1]/\sim$ is homeomorphic to $S^1$ and therefore compact, and we already know that $\Bbb R/\Bbb Z$ is Hausdorff, so we can use the result in my answer to conclude that the map is a homeomorphism. – Brian M. Scott Oct 23 '20 at 16:04
  • But if $\mathbb R/ \mathbb Z$ is Hausdorff, then $\pi \circ \iota$ is a quotient map $[0, 1]$ mod $\sim$? Or not? – PaulichenT Oct 23 '20 at 16:18
  • @PaulichenT: We don’t care whether $\pi\circ\iota$ is a quotient map; all that matters is that it is continuous and that it sends $0$ and $1$ to the point in $\Bbb R/\Bbb Z$. Then the universal property immediately gives us the continuous bijection from $[0,1]/\sim$ to $\Bbb R/\Bbb Z$. – Brian M. Scott Oct 23 '20 at 16:28
  • Ok, I got it. Thank you very much for the explanation. – PaulichenT Oct 23 '20 at 20:48
  • @PaulichenT: You’re very welcome. – Brian M. Scott Oct 23 '20 at 20:49